A) \[\frac{E}{R}\]
B) \[\frac{E}{7R}\]
C) \[\frac{2E}{7R}\]
D) \[\frac{2E}{R}\]
Correct Answer: C
Solution :
\[{{R}_{eq}}=R+\frac{4R}{3}=\frac{7R}{3}\] Current supplied by the battery \[=\frac{E}{{{R}_{eq}}}=\frac{3E}{7R}\] This current gets divided in the parallel branches in inverse ratio of branch resistances. Here, we will see that resistance of the other parallel branch is what fraction of the total resistance. The current in the required branch will also be the same fraction of the total current entering the node. So, \[{{I}_{2R}}=\frac{4}{6}\times \frac{3E}{7R}=\frac{2E}{7R}\]
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